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Common Core: High School - Geometry : Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 131 , $\angle B= 29 ^{\circ}$ , and $\angle C= 108 ^{\circ}$ .

Possible Answers:

$\\BC = 187.88 \\AC = 70.78 \\\angle A = 43^\circ$

$\\BC = 93.94 \\AC = 66.78 \\\angle A = 43^\circ$

$\\BC = 89.94 \\AC = 53.424 \\\angle A = 43^\circ$

$\\BC = 96.94 \\AC = 20.034 \\\angle A = 43^\circ$

$\\BC = 281.82 \\AC = 76.78 \\\angle A = 43^\circ$

Correct answer:

$\\BC = 93.94 \\AC = 66.78 \\\angle A = 43^\circ$

Explanation:

The first step is to create a picture

Plot5

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 29 - 108 \\\angle A = 43^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 29 for B , 108 for A and 131 for a , and solve for the missing side.

$\\\left(\frac{\sin( 108 )}{ 131 }\right)= \left(\frac{\sin( 29 )}{ AC }\right) \\\\AC = \left(\frac{ 131 \sin( 29 )}{\sin( 108 )}\right) \\\\AC = 66.7784292143583$

Remember to round to the nearest hundredth.

AC = 66.78

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 4459.5684 + 17161 - 2 \cdot 66.78 \cdot 131 \cdot \cos( 43 ) \\BC^{2} = 8824.54074913841 \\BC = 93.9390267627806 \\BC = 93.94$

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Example Question #22 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 116 , $\angle B= 27 ^{\circ}$ , and $\angle C= 120 ^{\circ}$ .

Possible Answers:

$\\BC = 68.95 \\AC = 48.648 \\\angle A = 33^\circ$

$\\BC = 75.95 \\AC = 18.243 \\\angle A = 33^\circ$

$\\BC = 218.85 \\AC = 70.81 \\\angle A = 33^\circ$

$\\BC = 75.95 \\AC = 18.243 \\\angle A = 33^\circ$

$\\BC = 218.85 \\AC = 70.81 \\\angle A = 33^\circ$

$\\BC = 68.95 \\AC = 48.648 \\\angle A = 33^\circ$

$\\BC = 72.95 \\AC = 60.81 \\\angle A = 33^\circ$

$\\BC = 145.9 \\AC = 64.81 \\\angle A = 33^\circ$

$\\BC = 72.95 \\AC = 60.81 \\\angle A = 33^\circ$

Correct answer:

$\\BC = 72.95 \\AC = 60.81 \\\angle A = 33^\circ$

Explanation:

The first step is to create a picture

Plot6

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 27 - 120 \\\angle A = 33^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 27 for B , 120 for A and 116 for a , and solve for the missing side.

$\\\left(\frac{\sin( 120 )}{ 116 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\\\AC = \left(\frac{ 116 \sin( 27 )}{\sin( 120 )}\right) \\\\AC = 60.8098766383251$

Remember to round to the nearest hundredth.

AC = 60.81

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 3697.8561000000004 + 13456 - 2 \cdot 60.81 \cdot 116 \cdot \cos( 33 ) \\BC^{2} = 5321.95882107139 \\BC = 72.9517568059289 \\BC = 72.95$

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Example Question #23 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 111 , $\angle B= 27 ^{\circ}$ , and $\angle C= 123 ^{\circ}$ .

Possible Answers:

$\\BC = 62.17 \\AC = 48.072 \\\angle A = 30^\circ$

$\\BC = 132.34 \\AC = 64.09 \\\angle A = 30^\circ$

$\\BC = 69.17 \\AC = 18.027 \\\angle A = 30^\circ$

$\\BC = 198.51 \\AC = 70.09 \\\angle A = 30^\circ$

$\\BC = 66.17 \\AC = 60.09 \\\angle A = 30^\circ$

Correct answer:

$\\BC = 66.17 \\AC = 60.09 \\\angle A = 30^\circ$

Explanation:

The first step is to create a picture

Plot7

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 27 - 123 \\\angle A = 30^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 27 for B , 123 for A and 111 for a , and solve for the missing side.

$\\\left(\frac{\sin( 123 )}{ 111 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\\\AC = \left(\frac{ 111 \sin( 27 )}{\sin( 123 )}\right) \\\\AC = 60.0866983976108$

Remember to round to the nearest hundredth.

AC = 60.09

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 3610.8081 + 12321 - 2 \cdot 60.09 \cdot 111 \cdot \cos( 30 ) \\BC^{2} = 4379.04653402366 \\BC = 66.174364628787 \\BC = 66.17$

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Example Question #24 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 100 , $\angle B= 15 ^{\circ}$ , and $\angle C= 97 ^{\circ}$ .

Possible Answers:

$\\BC = 89.41 \\AC = 20.864 \\\angle A = 68^\circ$

$\\BC = 93.41 \\AC = 26.08 \\\angle A = 68^\circ$

$\\BC = 280.23 \\AC = 36.08 \\\angle A = 68^\circ$

$\\BC = 96.41 \\AC = 7.824 \\\angle A = 68^\circ$

$\\BC = 186.82 \\AC = 30.08 \\\angle A = 68^\circ$

Correct answer:

$\\BC = 93.41 \\AC = 26.08 \\\angle A = 68^\circ$

Explanation:

The first step is to create a picture

Plot8

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 15 - 97 \\\angle A = 68^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 15 for B , 97 for A and 100 for a , and solve for the missing side.

$\\\left(\frac{\sin( 97 )}{ 100 }\right)= \left(\frac{\sin( 15 )}{ AC }\right) \\\\AC = \left(\frac{ 100 \sin( 15 )}{\sin( 97 )}\right) \\\\AC = 26.0762730956666$

Remember to round to the nearest hundredth.

AC = 26.08

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 680.1664 + 10000 - 2 \cdot 26.08 \cdot 100 \cdot \cos( 68 ) \\BC^{2} = 8726.2184087426 \\BC = 93.4142302261417 \\BC = 93.41$

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Example Question #25 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 109 , $\angle B= 16 ^{\circ}$ , and $\angle C= 43 ^{\circ}$ .

Possible Answers:

$\\BC = 132.99 \\AC = 35.24 \\\angle A = 121^\circ$

$\\BC = 139.99 \\AC = 13.215 \\\angle A = 121^\circ$

$\\BC = 136.99 \\AC = 44.05 \\\angle A = 121^\circ$

$\\BC = 410.97 \\AC = 54.05 \\\angle A = 121^\circ$

$\\BC = 273.98 \\AC = 48.05 \\\angle A = 121^\circ$

Correct answer:

$\\BC = 136.99 \\AC = 44.05 \\\angle A = 121^\circ$

Explanation:

The first step is to create a picture

Plot9

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 16 - 43 \\\angle A = 121^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 16 for B , 43 for A and 109 for a , and solve for the missing side.

$\\\left(\frac{\sin( 43 )}{ 109 }\right)= \left(\frac{\sin( 16 )}{ AC }\right) \\\\AC = \left(\frac{ 109 \sin( 16 )}{\sin( 43 )}\right) \\\\AC = 44.0535836204938$

Remember to round to the nearest hundredth.

AC = 44.05

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 1940.4024999999997 + 11881 - 2 \cdot 44.05 \cdot 109 \cdot \cos( 121 ) \\BC^{2} = 18767.2616295538 \\BC = 136.993655435402 \\BC = 136.99$

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Example Question #26 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 99 , $\angle B= 2 ^{\circ}$ , and $\angle C= 79 ^{\circ}$ .

Possible Answers:

$\\BC = 99.61 \\AC = 3.52 \\\angle A = 99^\circ$

$\\BC = 298.83 \\AC = 13.52 \\\angle A = 99^\circ$

$\\BC = 102.61 \\AC = 1.056 \\\angle A = 99^\circ$

$\\BC = 95.61 \\AC = 2.816 \\\angle A = 99^\circ$

$\\BC = 199.22 \\AC = 7.52 \\\angle A = 99^\circ$

Correct answer:

$\\BC = 99.61 \\AC = 3.52 \\\angle A = 99^\circ$

Explanation:

The first step is to create a picture

Plot10

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 2 - 79 \\\angle A = 99^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 2 for B , 79 for A and 99 for a , and solve for the missing side.

$\\\left(\frac{\sin( 79 )}{ 99 }\right)= \left(\frac{\sin( 2 )}{ AC }\right) \\\\AC = \left(\frac{ 99 \sin( 2 )}{\sin( 79 )}\right) \\\\AC = 3.51971729370085$

Remember to round to the nearest hundredth.

AC = 3.52

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 12.3904 + 9801 - 2 \cdot 3.52 \cdot 99 \cdot \cos( 99 ) \\BC^{2} = 9922.41896475444 \\BC = 99.6113395390025 \\BC = 99.61$

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Example Question #27 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 47 , $\angle B= 16 ^{\circ}$ , and $\angle C= 78 ^{\circ}$ .

Possible Answers:

$\\BC = 47.93 \\AC = 13.24 \\\angle A = 86^\circ$

$\\BC = 143.79 \\AC = 23.24 \\\angle A = 86^\circ$

$\\BC = 95.86 \\AC = 17.24 \\\angle A = 86^\circ$

$\\BC = 50.93 \\AC = 3.972 \\\angle A = 86^\circ$

$\\BC = 43.93 \\AC = 10.592 \\\angle A = 86^\circ$

Correct answer:

$\\BC = 47.93 \\AC = 13.24 \\\angle A = 86^\circ$

Explanation:

The first step is to create a picture

Plot11

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 16 - 78 \\\angle A = 86^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 16 for B , 78 for A and 47 for a , and solve for the missing side.

$\\\left(\frac{\sin( 78 )}{ 47 }\right)= \left(\frac{\sin( 16 )}{ AC }\right) \\\\AC = \left(\frac{ 47 \sin( 16 )}{\sin( 78 )}\right) \\\\AC = 13.2443771407098$

Remember to round to the nearest hundredth.

AC = 13.24

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 175.29760000000002 + 2209 - 2 \cdot 13.24 \cdot 47 \cdot \cos( 86 ) \\BC^{2} = 2297.48148303701 \\BC = 47.9320506867483 \\BC = 47.93$

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Example Question #28 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 187 , $\angle B= 29 ^{\circ}$ , and $\angle C= 55 ^{\circ}$ .

Possible Answers:

$\\BC = 230.03 \\AC = 33.201 \\\angle A = 96^\circ$

$\\BC = 681.09 \\AC = 120.67 \\\angle A = 96^\circ$

$\\BC = 454.06 \\AC = 114.67 \\\angle A = 96^\circ$

$\\BC = 223.03 \\AC = 88.536 \\\angle A = 96^\circ$

$\\BC = 227.03 \\AC = 110.67 \\\angle A = 96^\circ$

Correct answer:

$\\BC = 227.03 \\AC = 110.67 \\\angle A = 96^\circ$

Explanation:

The first step is to create a picture

Plot12

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 29 - 55 \\\angle A = 96^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

$\frac{1}{a} \sin{\left (A \right )} = \frac{1}{AC} \sin{\left (B \right )}$

Now plug in 29 for B , 55 for A and 187 for a , and solve for the missing side.

$\\\left(\frac{\sin( 55 )}{ 187 }\right)= \left(\frac{\sin( 29 )}{ AC }\right) \\\\AC = \left(\frac{ 187 \sin( 29 )}{\sin( 55 )}\right) \\\\AC = 110.674690514574$

Remember to round to the nearest hundredth.

AC = 110.67

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 12247.8489 + 34969 - 2 \cdot 110.67 \cdot 187 \cdot \cos( 96 ) \\BC^{2} = 51543.3426211569 \\BC = 227.03158947855 \\BC = 227.03$

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Example Question #93 : Similarity, Right Triangles, & Trigonometry

Using the Law of Cosines, solve for . Round to the second decimal place.

Screen shot 2020 08 07 at 2.49.50 pm

Possible Answers:

x = 45

x = 15.13

x = 28.96

x = 67.89

Correct answer:

x = 28.96

Explanation:

We will begin by defining the variables in our equation $A^2 + B^2 -2ABcos(\gamma) = C^2$ .

We will let

A = 16

B = 12

C = 8

$\gamma = x$

Next we plug our values into our formulas and solve for .

$A^2 + B^2 -2ABcos(\gamma) = C^2$

$\implies 16^2 + 12^2 - 2(16)(12)cos(x) = 8^2$

$\implies 256 + 144 - 384cos(x) = 64$

$\implies -384cos(x) = -336$

$\implies cos(x) = 0.875$

$\implies x = cos^{-1}(0.875)$

$\implies x = 28.96$

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Example Question #21 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Which of the following expressions properly states the Law of Cosines? Use the following triangle for reference.

Screen shot 2020 08 13 at 8.30.42 am

Possible Answers:

$A^2 + B^2 -2ABcos(\gamma) = C^2$

$\frac{A}{cos(\gamma)} = \frac{B}{cos(\alpha)} = \frac{C}{cos(\beta)}$

$A^2 + B^2 -2ABsin(\gamma) = C^2$

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

Correct answer:

$A^2 + B^2 -2ABsin(\gamma) = C^2$

Explanation:

The Law of Cosines should look similar to a theorem we use quite commonly for right triangles. The Law of Cosines is basically the Pythagorean Theorem, but it has an added argument so that it is true for any and all triangles.

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #21 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #22 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #23 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #24 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #25 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #26 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #27 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #28 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #93 : Similarity, Right Triangles, & Trigonometry

Example Question #21 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

All Common Core: High School - Geometry Resources

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